// 想定解法: DP 全パターン生成 モンテカルロ

#include <iostream>
#include <cstring>
#include <iomanip>
using namespace std;
typedef long long ll;
#define REP(i, n)           for(int(i)=0;(i)<(n);++(i))
#define RREP(i, n)          for(int(i)=(n)-1;(i)>=0;--(i))
void rc(int v,int mn,int mx){if(v<mn||mx<v){cerr<<"error"<<endl;}}

const int MAX = 10;
const int MAXV = MAX*6;
ll dp1[MAXV+1];
ll dp2[MAXV+1];

int main(){
    int N, K;
    cin >> N >> K;
    rc(N,1,MAX);
    rc(K,0,N);

    REP(i,MAXV+1) dp1[i] = 0;
    dp1[0] = 1;
    REP(i,N-K) RREP(j,MAXV+1){
        REP(k,6) dp1[j+k+1] += dp1[j]; dp1[j] = 0;
    }

    memcpy(dp2, dp1, sizeof(dp2));

    REP(i,K) RREP(j,MAXV+1){
        REP(k,6) dp1[j+k+1] += dp1[j]; dp1[j] = 0;
        REP(k,3) dp2[j+k+4] += dp2[j]*2; dp2[j] = 0;
    }

    REP(j,MAXV) dp1[j+1] += dp1[j];
    ll sum = dp1[MAXV];
    ll win = 0, total = 0;

    REP(i,MAXV+1){
        if(!dp2[i]) continue;
        win += dp1[i-1] * dp2[i];
        total += sum * dp2[i];
    }
    cout << fixed << setprecision(15) << 1.0*win/total << endl;
    return 0;
}